Question

Consider a 640 mL solution containing 33.4 g of NaF and a 640 mL solution containing 210.1 g of BaF2. What are the molar concentrations of sodium, barium, and fluoride ions when the two solutions are mixed together?

Answer 1

1 mol equivalent NaF = ( 23 + 19 ) g of NaF = 42 g of NaF

1000 mL of solution containing 42 g of NaF = 1 (M) solution

1mL of solution containing 1 g of NaF = 1 (M) * /(42 * 1000)

640 mL of solution containing 33.4 g of NaF= ( 1(M) * 640 * 33.4 ) / ( 42 * 1000)

= 0.534 ( M)

Similarly,  

640 mL of solution containing 210.1 g of BaF₂ = ( 1 (M) * 210.1 * 640 ) / ( 44 * 1000 )

= 3.056 (M)

When NaF and BaF₂ samples of 640 mL each are mixed then the mixed solution became 640+ 640 = 1280 mL of solution.

In 1280 mL of solution there are 1 ion of Na⁺ , 1 ion of Ba²⁺ , and 3 ions of F^-1 are present.

42 g of NaF contain in 1000 mL solution = 23 g of Na

1g of NaF contain in 1 mL solution = 23/ ( 42 * 1000) g of Na

33.4 g of NaF contain 640 mL solution =( 23 * 33.4 * 640 ) / ( 42 * 1000) = 11.70 g of Na

In mixer solution ,

11.70 g of Na⁺ ion present in 1280 mL of mixed solution .

Molar concentration of Na⁺ in mixed solution = ( 1 (M) * 11.70 *1280 ) / ( 23 * 1000 ) = 0.651 (M)

The Moleculer weight of BaF₂ = 6 + 2*19 = 44

44 g of BaF₂ contain in 1000 mL of solution =  6 g of Ba

1g of BaF₂ contain in 1 mL of solution = 6 / (44* 1000) g of Ba

210.1g of BaF₂ contain 640 mL of solution =( 6 * 210.1 * 640 ) / ( 44 * 1000 ) g of Ba = 18.336 g of Ba

18.336 g of Ba²⁺ ion present in 1280 of mixed solution .

Molar concentration of Ba²⁺ mixed solution =( 1 (M) * 18.336 * 1280 ) / ( 1000* 210.1 ) = 0.111 (M )

In 33.4 g of NaF = ( 33.4 - 11.70) = 21.7 g of F

In 210.1 g of BaF₂ = ( 210.1 - 18.336 ) = 191.764 g of F^- ion

In mixed solution ( 21.7 + 191.764 ) = 213.464 g of F^- ion

Total ( 210.1 + 33.4) = 243.5 g of mixture 213.464 g of F^- ion

Molar concentration of F^- ion =( 1 (M) * 213.464 * 1280) / ( 1000 * 243.5 ) = 1.122 (M)